Brightstorm is like having a personal tutor for every subject
See what all the buzz is aboutCheck it out
Finding the Maximum or Minimum of a Quadratic - Problem 2 2,456 views
So whenever we're given an application of a quadratic equation, one of the things we can do is look for their vertex to find the maximum or minimum of whatever they are giving us.
So often this is used as a ball being tossed up and down or something like that, but there can be other uses as well. So for this particular problem what we're looking at is an area and what we're looking at is a town is fencing in a playground which is bordered on one side of a building. The plan is to use 640 feet of fence on the other three sides, what should the dimensions be if the playground is to be maximized.?
So what we have is a building, so you obviously don't need to fence that side and then the other three sides are bordered by this fence of which we have 640 feet of fencing. So I'm going to take this information and go ahead and make a picture in hopes of somehow making an equation of it.
So what we have is a building and then we have this fence and we know that it's going to be a rectangle, so we have this building no fences are needed there and we have this field that we're trying to make. We have 640 feet of fence altogether. We don't know what each side is going to be, but what we do know is that these two sides are going to be the same, it's a rectangle the two sides are going to be the same, so it's easiest to call those two sides x, x and x.
So if I've used up x and x, how much does that leave me for the third side. Think about if we use say just an easier number 100 feet and each of these sides was 10, we would have used up 10 for each side 20 so we would have had 100 minus 20 or 80. So we're using both of these which leaves this side left as 640 minus 2x.
So now we want a equation for the area; area is just length times width. So the area of x is just going to be x times 640 minus 2x. Foiling this out, what we end up with is ax is equal to -2xÂ² plus 640x.
Now we're looking to maximize this area which means we want to find new vertex because the vertex is going to be at the highest point. I know this is a downward facing parabola, so therefore the largest point is going to be at the top at the vertex. The easiest way to find the vertex is -b over 2a which leaves me with -640 over -4, 640 divided by 4 is 160.
So what we found is that our width then is 160 and we're asked to just find the dimensions of the largest rectangle, so if the largest width is 160, plug in 160 to find the other side, 2 times 160 is 320 640 minus 320 is 320, so what we were able to find is that the other side is 320 feet giving our dimensions 160 by 320.
So taking a word problem, making a picture and then trying to optimize that trying to find the highest or lowest point by finding the vertex.